What is the extraneous solution to these equations? $\dfrac{x^2 + 9}{x - 3} = \dfrac{17x - 63}{x - 3}$
Multiply both sides by $x - 3$ $ \dfrac{x^2 + 9}{x - 3} (x - 3) = \dfrac{17x - 63}{x - 3} (x - 3)$ $ x^2 + 9 = 17x - 63$ Subtract $17x - 63$ from both sides: $ x^2 + 9 - (17x - 63) = 17x - 63 - (17x - 63)$ $ x^2 + 9 - 17x + 63 = 0$ $ x^2 + 72 - 17x = 0$ Factor the expression: $ (x - 8)(x - 9) = 0$ Therefore $x = 8$ or $x = 9$ The original expression is defined at $x = 8$ and $x = 9$, so there are no extraneous solutions.